Finite covers of $X$ and continuous profinite group actions

Question

I have a question that is related to the one here, but I found the answer there unsatisfactory and kind of confusing.

For context, we are considering the action of the fundamental group $\pi_1(X,x)$ on the fiber $p^{-1}(x)$ of some finite connected cover $p:Y\rightarrow X$. The proof is from Corollary 2.3.9 of Szamuely's book, showing an equivalence of categories of finite covers of $X$ with the category of finite continuous left $\hat{\pi}_1(X,x)$-sets, where $\hat{\pi}_1(X,x)$ is the profinite completion of $\pi_1(X,x)$.

Here's the result from the text:

For a finite connected cover $p\colon Y\rightarrow X$ the action of $\pi_1 (X,x)$ on $p^{-1}(x)$ factors through a finite quotient, so we obtain an action of $\hat{\pi}_1 (X,x)$ as well. The stabilizer of each point $y\in Y$ [$y\in p^{-1}(x)$] is a subgroup of finite index, and hence contains a normal subgroup of finite index by the lemma. Therefore the stabilizer of $y$ under the action of $\hat{\pi}_1 (X,x)$ is an open subgroup in the profinite topology (being a union of cosets of an open normal subgroup), which means that the action is continuous. Conversely, a continuous action of $\hat{\pi}_1(X,x)$ on a finite set factors through a finite quotient which is also a quotient of $\pi_1(X,x)$, and as such gives rise to a finite cover $Y\rightarrow X$.

I guess I'm mainly confused on two points here:

1. Why is the normal subgroup open? I tried drawing some diagrams and couldn't come up with a satisfactory answer. The answer I linked above says that we can see that the projection $\hat{\pi}_1(X,x)\rightarrow \hat{\pi}_1(X,x)/N$ is continuous where the image has the discrete topology, but it seems to me that the argument then implies that any normal $N$ with finite index in a profinite group would be open, which I don't think is true.

2. It also is not clear to me why a finite quotient of $\hat{\pi}_1(X,x)$ is also a finite quotient of $\pi_1(X,x)$. I tried messing around with the natural map $\pi_1(X,x)\rightarrow \hat{\pi}_1(X,x)$ but didn't get anywhere.

Answer 1

Recall that the pro-finite completion $\widehat G$ of a group $G$ is the limit of $G/N$ where $N$ ranges over all finite-index normal subgroups of $G$. Thus, any open normal subgroup $N'$ of $G$ gives rise to an open subgroup $\widehat N\subset \widehat G$, since $\widehat N$ is the limit of $N/(N\cap N')\simeq NN'/N$ where $N'\subset G$ is a normal subgroup of finite index. It comes with a natural isomorphism $G/N\simeq \widehat G/\widehat N$.

The point is that the action of $\pi_1(X,x)$ on $\pi^{-1}(x)$ is the same as a homomorphism $\pi_1(X,x)\to S_n$. Let $N$ be the kernel, which is a normal subgroup of finite index in $\pi_1(X,x)$. Let $\widehat N$ be the kernel of the extended homomorphism $\widehat\pi_1(X,x)\to S_n$. Then by the above discussion, we have an isomorphism $\pi_1/N\simeq\widehat \pi_1/\widehat N$.

Now the stabilizer in $\widehat\pi_1(X,x)$ of a point $y\in p^{-1}(x)$ is some group $\widehat N\subset S\subset \widehat\pi_1(X,x)$. Thus $S$ is a union of translates of $\widehat N$, hence is also open. Moreover, it is finite index since $\widehat N$ is already finite-index in $\widehat\pi_1(X,x)$.

To address question 2, the claim is that for any group $G$, a finite-index open subgroup $M$ of $\widehat G$ arises from as $\widehat N$ for some finite-index normal subgroup $N$ of $G$. The point is that $N$ is simply the kernel of the composition $G\to\widehat G\to\widehat G/M$.